Poisson bracket#Hamilton's equations of motion

Given two functions {{mvar|f}} and {{mvar|g}} that depend on phase space and time, their Poisson bracket $\backslash \{f,\; g\backslash \}$ is another function that depends on phase space and time. The following rules hold for any three functions $f,\backslash ,\; g,\backslash ,\; h$ of phase space and time:

;Anticommutativity: $\backslash \{f,\; g\backslash \}\; =\; -\backslash \{g,\; f\backslash \}$

;Bilinearity: $\backslash \{af\; +\; bg,\; h\backslash \}\; =\; a\backslash \{f,\; h\backslash \}\; +\; b\backslash \{g,\; h\backslash \},$$\backslash \{h,\; af\; +\; bg\backslash \}\; =\; a\backslash \{h,\; f\backslash \}\; +\; b\backslash \{h,\; g\backslash \},\; \backslash quad\; a,\; b\; \backslash in\; \backslash mathbb\; R$

;Leibniz's rule: $\backslash \{fg,\; h\backslash \}\; =\; \backslash \{f,\; h\backslash \}g\; +\; f\backslash \{g,\; h\backslash \}$

;Jacobi identity: $\backslash \{f,\; \backslash \{g,\; h\backslash \}\backslash \}\; +\; \backslash \{g,\; \backslash \{h,\; f\backslash \}\backslash \}\; +\; \backslash \{h,\; \backslash \{f,\; g\backslash \}\backslash \}\; =\; 0$

Also, if a function $k$ is constant over phase space (but may depend on time), then $\backslash \{f,\backslash ,\; k\backslash \}\; =\; 0$ for any $f$.

In canonical coordinates (also known as Darboux coordinates) $(q\_i,\backslash ,\; p\_i)$ on the phase space, given two functions $f(p\_i,\backslash ,\; q\_i,\; t)$ and $g(p\_i,\backslash ,\; q\_i,\; t)$,$f(p\_i,\backslash ,\; q\_i,\backslash ,\; t)$ means $f$ is a function of the $2N\; +\; 1$ independent variables: momentum, $p\_\{1\; \backslash dots\; N\}$; position, $q\_\{1\; \backslash dots\; N\}$; and time, $t$ the Poisson bracket takes the form

$$\backslash \{f,\; g\backslash \}\; =\; \backslash sum\_\{i=1\}^\{N\}\; \backslash left(\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; q\_\{i\}\}\; \backslash frac\{\backslash partial\; g\}\{\backslash partial\; p\_\{i\}\}\; -\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; p\_i\}\; \backslash frac\{\backslash partial\; g\}\{\backslash partial\; q\_i\}\backslash right).$$

The Poisson brackets of the canonical coordinates are

$$\backslash begin\{align\}$$

\{q_k,q_l\} &= \sum_{i=1}^{N} \left( \frac{\partial q_k}{\partial q_{i}} \frac{\partial q_l}{\partial p_{i}} - \frac{\partial q_k}{\partial p_i} \frac{\partial q_l}{\partial q_i}\right) = \sum_{i=1}^{N} \left( \delta_{ki} \cdot 0 - 0 \cdot \delta_{li}\right) = 0, \\

\{p_k,p_l\} &=\sum_{i=1}^{N} \left( \frac{\partial p_k}{\partial q_{i}} \frac{\partial p_l}{\partial p_{i}} - \frac{\partial p_k}{\partial p_i} \frac{\partial p_l}{\partial q_i}\right) = \sum_{i=1}^{N} \left( 0 \cdot \delta_{li} - \delta_{ki} \cdot 0\right) = 0, \\

\{q_k,p_l\} &= \sum_{i=1}^{N} \left( \frac{\partial q_k}{\partial q_{i}} \frac{\partial p_l}{\partial p_{i}} - \frac{\partial q_k}{\partial p_i} \frac{\partial p_l}{\partial q_i}\right) = \sum_{i=1}^{N} \left( \delta_{ki} \cdot \delta_{li} - 0 \cdot 0\right) = \delta_{kl},

\end{align}

where $\backslash delta\_\{ij\}$ is the Kronecker delta.

Hamilton's equations of motion have an equivalent expression in terms of the Poisson bracket. This may be most directly demonstrated in an explicit coordinate frame. Suppose that $f(p,\; q,\; t)$ is a function on the solution's trajectory-manifold. Then from the multivariable chain rule,

$$\backslash frac\{d\}\{dt\}\; f(p,\; q,\; t)\; =\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; q\}\; \backslash frac\{dq\}\{dt\}\; +\; \backslash frac\; \{\backslash partial\; f\}\{\backslash partial\; p\}\; \backslash frac\{dp\}\{dt\}\; +\; \backslash frac\{\backslash partial\; f\}\{\backslash partial\; t\}.$$

Further, one may take $p\; =\; p(t)$ and $q\; =\; q(t)$ to be solutions to Hamilton's equations; that is,

$$\backslash begin\{align\}$$

\frac{d q}{d t} &= \frac{\partial \mathcal H}{\partial p} = \{q, \mathcal H\}, \\

\frac{d p}{d t} &= -\frac{\partial \mathcal H}{\partial q} = \{p, \mathcal H\}.

\end{align}

Then

$$\backslash begin\{align\}$$

\frac {d}{dt} f(p, q, t) &= \frac{\partial f}{\partial q} \frac{\partial \mathcal H}{\partial p} - \frac{\partial f}{\partial p} \frac{\partial \mathcal H}{\partial q} + \frac{\partial f}{\partial t} \\

&= \{f, \mathcal H\} + \frac{\partial f}{\partial t} ~.

\end{align}

Thus, the time evolution of a function $f$ on a symplectic manifold can be given as a one-parameter family of symplectomorphisms (i.e., canonical transformations, area-preserving diffeomorphisms), with the time $t$ being the parameter: Hamiltonian motion is a canonical transformation generated by the Hamiltonian. That is, Poisson brackets are preserved in it, so that any time $t$ in the solution to Hamilton's equations,

$$q(t)\; =\; \backslash exp\; (-t\; \backslash \{\; \backslash mathcal\; H,\; \backslash cdot\; \backslash \}\; )\; q(0),\; \backslash quad\; p(t)\; =\; \backslash exp\; (-t\; \backslash \{\; \backslash mathcal\; H,\; \backslash cdot\; \backslash \})\; p(0),$$

can serve as the bracket coordinates. Poisson brackets are canonical invariants.

Dropping the coordinates,

$$\backslash frac\{d\}\{dt\}\; f\; =\; \backslash left(\backslash frac\{\backslash partial\}\{\backslash partial\; t\}\; -\; \backslash \{\backslash mathcal\; H,\; \backslash cdot\backslash \}\backslash right)f.$$

The operator in the convective part of the derivative, $i\backslash hat\{L\}\; =\; -\backslash \{\backslash mathcal\; H,\; \backslash cdot\backslash \}$, is sometimes referred to as the Liouvillian (see Liouville's theorem (Hamiltonian)).

{{Main|Canonical transformation}}

The concept of Poisson brackets can be expanded to that of matrices by defining the Poisson matrix.

Consider the following canonical transformation:$$\backslash eta\; =$$

\begin{bmatrix}

q_1\\

\vdots \\

q_N\\

p_1\\

\vdots\\

p_N\\

\end{bmatrix} \quad \rightarrow \quad \varepsilon =

\begin{bmatrix}

Q_1\\

\vdots \\

Q_N\\

P_1\\

\vdots\\

P_N\\

\end{bmatrix} Defining $M\; :=\; \backslash frac\{\backslash partial\; (\backslash mathbf\{Q\},\; \backslash mathbf\{P\})\}\{\backslash partial\; (\backslash mathbf\{q\},\; \backslash mathbf\{p\})\}$, the Poisson matrix is defined as $\backslash mathcal\; P(\backslash varepsilon)\; =\; MJM^T$

, where $J$ is the symplectic matrix under the same conventions used to order the set of coordinates. It follows from the definition that:$$\backslash mathcal\; P\_\{ij\}(\backslash varepsilon)\; =\; [MJM^T]\_\{ij\}=\backslash sum\_\{k=1\}^\{N\}\; \backslash left(\; \backslash frac\{\backslash partial\; \backslash varepsilon\_i\}\{\backslash partial\; \backslash eta\_\{k\}\}\; \backslash frac\{\backslash partial\; \backslash varepsilon\_j\}\{\backslash partial\; \backslash eta\_\{N+k\}\}\; -\; \backslash frac\{\backslash partial\; \backslash varepsilon\_i\}\{\backslash partial\; \backslash eta\_\{N+k\}\}\; \backslash frac\{\backslash partial\; \backslash varepsilon\_j\}\{\backslash partial\; \backslash eta\_k\}\backslash right)=\backslash sum\_\{k=1\}^\{N\}\; \backslash left(\; \backslash frac\{\backslash partial\; \backslash varepsilon\_i\}\{\backslash partial\; q\_\{k\}\}\; \backslash frac\{\backslash partial\; \backslash varepsilon\_j\}\{\backslash partial\; p\_k\}\; -\; \backslash frac\{\backslash partial\; \backslash varepsilon\_i\}\{\backslash partial\; p\_k\}\; \backslash frac\{\backslash partial\; \backslash varepsilon\_j\}\{\backslash partial\; q\_k\}\backslash right)=\backslash \{\; \backslash varepsilon\_i,\backslash varepsilon\_j\backslash \}\_\backslash eta.$$

The Poisson matrix satisfies the following known properties:$$\backslash begin\{align\}$$

\mathcal P^T &= - \mathcal P \\

|\mathcal P| &= \frac{1}{|M|^2}\\

\mathcal P^{-1}(\varepsilon)&= -(M^{-1})^T J M^{-1} = - \mathcal L (\varepsilon)\\

\end{align}

where the $\backslash mathcal\; L(\backslash varepsilon)$

is known as a Lagrange matrix and whose elements correspond to Lagrange brackets. The last identity can also be stated as the following:$$\backslash sum\_\{k=1\}^\{2N\}\; \backslash \{\backslash eta\_i,\backslash eta\_k\backslash \}[\backslash eta\_k,\backslash eta\_j]\; =\; -\backslash delta\_\{ij\}$$

Note that the summation here involves generalized coordinates as well as generalized momentum.

The invariance of Poisson bracket can be expressed as: $\backslash \{\; \backslash varepsilon\_i,\backslash varepsilon\_j\backslash \}\_\backslash eta=\backslash \{\; \backslash varepsilon\_i,\backslash varepsilon\_j\backslash \}\_\backslash varepsilon\; =\; J\_\{ij\}$

, which directly leads to the symplectic condition: $MJM^T\; =\; J$

.{{Cite book |last=Giacaglia |first=Giorgio E. O. |title=Perturbation methods in non-linear systems |date=1972 |publisher=Springer |isbn=978-3-540-90054-2 |series=Applied mathematical sciences |location=New York Heidelberg |pages=8–9}}

An integrable system will have constants of motion in addition to the energy. Such constants of motion will commute with the Hamiltonian under the Poisson bracket. Suppose some function $f(p,\; q)$ is a constant of motion. This implies that if $p(t),\; q(t)$ is a trajectory or solution to Hamilton's equations of motion, then along that trajectory:$$0\; =\; \backslash frac\{df\}\{dt\}$$Where, as above, the intermediate step follows by applying the equations of motion and we assume that $f$ does not explicitly depend on time. This equation is known as the Liouville equation. The content of Liouville's theorem is that the time evolution of a measure given by a distribution function $f$ is given by the above equation.

If the Poisson bracket of $f$ and $g$ vanishes ($\backslash \{f,g\backslash \}\; =\; 0$), then $f$ and $g$ are said to be in involution. In order for a Hamiltonian system to be completely integrable, $n$ independent constants of motion must be in mutual involution, where $n$ is the number of degrees of freedom.

Furthermore, according to Poisson's Theorem, if two quantities $A$ and $B$ are explicitly time independent ($A(p,\; q),\; B(p,\; q)$) constants of motion, so is their Poisson bracket $\backslash \{A,\backslash ,\; B\backslash \}$. This does not always supply a useful result, however, since the number of possible constants of motion is limited ($2n\; -\; 1$ for a system with $n$ degrees of freedom), and so the result may be trivial (a constant, or a function of $A$ and $B$.)

Let $M$ be a symplectic manifold, that is, a manifold equipped with a symplectic form: a 2-form $\backslash omega$ which is both closed (i.e., its exterior derivative $d\; \backslash omega$ vanishes) and non-degenerate. For example, in the treatment above, take $M$ to be $\backslash mathbb\{R\}^\{2n\}$ and take

$$\backslash omega\; =\; \backslash sum\_\{i=1\}^\{n\}\; d\; q\_i\; \backslash wedge\; d\; p\_i.$$

If $\backslash iota\_v\; \backslash omega$ is the interior product or contraction operation defined by $(\backslash iota\_v\; \backslash omega)(u)\; =\; \backslash omega(v,\backslash ,\; u)$, then non-degeneracy is equivalent to saying that for every one-form $\backslash alpha$ there is a unique vector field $\backslash Omega\_\backslash alpha$ such that $\backslash iota\_\{\backslash Omega\_\backslash alpha\}\; \backslash omega\; =\; \backslash alpha$. Alternatively, $\backslash Omega\_\{d\; H\}\; =\; \backslash omega^\{-1\}(d\; H)$. Then if $H$ is a smooth function on $M$, the Hamiltonian vector field $X\_H$ can be defined to be $\backslash Omega\_\{d\; H\}$. It is easy to see that

$$\backslash begin\{align\}$$

X_{p_i} &= \frac{\partial}{\partial q_i} \\

X_{q_i} &= -\frac{\partial}{\partial p_i}.

\end{align}

The Poisson bracket $\backslash \; \backslash \{\backslash cdot,\backslash ,\; \backslash cdot\backslash \}$ on {{math|(M, ω)}} is a bilinear operation on differentiable functions, defined by $\backslash \{f,\backslash ,\; g\backslash \}\; \backslash ;=\backslash ;\; \backslash omega(X\_f,\backslash ,\; X\_g)$; the Poisson bracket of two functions on {{math|M}} is itself a function on {{math|M}}. The Poisson bracket is antisymmetric because:

$$\backslash \{f,\; g\backslash \}\; =\; \backslash omega(X\_f,\; X\_g)\; =\; -\backslash omega(X\_g,\; X\_f)\; =\; -\backslash \{g,\; f\backslash \}\; .$$

Furthermore,

{{NumBlk||$$\backslash begin\{align\}$$

\{f, g\} &= \omega(X_f, X_g) = \omega(\Omega_{df}, X_g) \\

&= (\iota_{\Omega_{df}}\omega)(X_g) = df(X_g) \\

&= X_g f = \mathcal{L}_{X_g} f.

\end{align}|{{EquationRef|1}}}}

Here {{math|X_gf}} denotes the vector field {{math|X_g}} applied to the function {{math|f}} as a directional derivative, and $\backslash mathcal\{L\}\_\{X\_g\}\; f$ denotes the (entirely equivalent) Lie derivative of the function {{math|f}}.

If {{math|α}} is an arbitrary one-form on {{math|M}}, the vector field {{math|Ω_α}} generates (at least locally) a flow $\backslash phi\_x(t)$ satisfying the boundary condition $\backslash phi\_x(0)\; =\; x$ and the first-order differential equation

$$\backslash frac\{d\backslash phi\_x\}\{dt\}\; =\; \backslash left.\; \backslash Omega\_\backslash alpha\; \backslash right|\_\{\backslash phi\_x(t)\}.$$

The $\backslash phi\_x(t)$ will be symplectomorphisms (canonical transformations) for every {{math|t}} as a function of {{math|x}} if and only if $\backslash mathcal\{L\}\_\{\backslash Omega\_\backslash alpha\}\backslash omega\; \backslash ;=\backslash ;\; 0$; when this is true, {{math|Ω_α}} is called a symplectic vector field. Recalling Cartan's identity $\backslash mathcal\{L\}\_X\backslash omega\; \backslash ;=\backslash ;\; d\; (\backslash iota\_X\; \backslash omega)\; \backslash ,+\backslash ,\; \backslash iota\_X\; d\backslash omega$ and {{math|1=dω = 0}}, it follows that $\backslash mathcal\{L\}\_\{\backslash Omega\_\backslash alpha\}\backslash omega\; \backslash ;=\backslash ;\; d\backslash left(\backslash iota\_\{\backslash Omega\_\backslash alpha\}\; \backslash omega\backslash right)\; \backslash ;=\backslash ;\; d\backslash alpha$. Therefore, {{math|Ω_α}} is a symplectic vector field if and only if α is a closed form. Since $d(df)\; \backslash ;=\backslash ;\; d^2f\; \backslash ;=\backslash ;\; 0$, it follows that every Hamiltonian vector field {{math|X_f}} is a symplectic vector field, and that the Hamiltonian flow consists of canonical transformations. From {{EquationNote|1|(1)}} above, under the Hamiltonian flow $X\_\backslash mathcal\; H$,

$$\backslash frac\{d\}\{dt\}f(\backslash phi\_x(t))\; =\; X\_\backslash mathcal\{H\}f\; =\; \backslash \{f,\backslash mathcal\; H\backslash \}.$$

This is a fundamental result in Hamiltonian mechanics, governing the time evolution of functions defined on phase space. As noted above, when {{math|1={f,\mathcal H} = 0}}, {{math|f}} is a constant of motion of the system. In addition, in canonical coordinates (with $\backslash \{p\_i,\backslash ,\; p\_j\backslash \}\; \backslash ;=\backslash ;\; \backslash \{q\_i,q\_j\backslash \}\; \backslash ;=\backslash ;\; 0$ and $\backslash \{q\_i,\backslash ,\; p\_j\backslash \}\; \backslash ;=\backslash ;\; \backslash delta\_\{ij\}$), Hamilton's equations for the time evolution of the system follow immediately from this formula.

It also follows from {{EquationNote|1|(1)}} that the Poisson bracket is a derivation; that is, it satisfies a non-commutative version of Leibniz's product rule:

{{NumBlk||$$\backslash \{fg,h\backslash \}\; =\; f\backslash \{g,h\backslash \}\; +\; g\backslash \{f,h\backslash \},$$ and $$\backslash \{f,gh\backslash \}\; =\; g\backslash \{f,h\backslash \}\; +\; h\backslash \{f,g\backslash \}.$$|{{EquationRef|2}}}}

The Poisson bracket is intimately connected to the Lie bracket of the Hamiltonian vector fields. Because the Lie derivative is a derivation,

$$\backslash mathcal\; L\_v\backslash iota\_u\backslash omega\; =\; \backslash iota\_\{\backslash mathcal\; L\_vu\}\backslash omega\; +\; \backslash iota\_u\backslash mathcal\; L\_v\backslash omega\; =\; \backslash iota\_\{[v,u]\}\backslash omega\; +\; \backslash iota\_u\backslash mathcal\; L\_v\backslash omega.$$

Thus if {{math|v}} and {{math|u}} are symplectic, using $\backslash mathcal\{L\}\_v\backslash omega\; =0=\backslash mathcal\; L\_u\backslash omega$, Cartan's identity, and the fact that $\backslash iota\_u\backslash omega$ is a closed form,

$$\backslash iota\_\{[v,u]\}\backslash omega\; =\; \backslash mathcal\; L\_v\backslash iota\_u\backslash omega\; =\; d(\backslash iota\_v\backslash iota\_u\backslash omega)\; +\; \backslash iota\_vd(\backslash iota\_u\backslash omega)\; =\; d(\backslash iota\_v\backslash iota\_u\backslash omega)\; =\; d(\backslash omega(u,v)).$$

It follows that $[v,u]\; =\; X\_\{\backslash omega(u,v)\}$, so that

{{NumBlk||$$[X\_f,X\_g]\; =\; X\_\{\backslash omega(X\_g,X\_f)\}\; =\; -X\_\{\backslash omega(X\_f,X\_g)\}\; =\; -X\_\{\backslash \{f,g\backslash \}\}.$$|{{EquationRef|3}}}}

Thus, the Poisson bracket on functions corresponds to the Lie bracket of the associated Hamiltonian vector fields. We have also shown that the Lie bracket of two symplectic vector fields is a Hamiltonian vector field and hence is also symplectic. In the language of abstract algebra, the symplectic vector fields form a subalgebra of the Lie algebra of smooth vector fields on {{math|M}}, and the Hamiltonian vector fields form an ideal of this subalgebra. The symplectic vector fields are the Lie algebra of the (infinite-dimensional) Lie group of symplectomorphisms of {{math|M}}.

It is widely asserted that the Jacobi identity for the Poisson bracket,

$$\backslash \{f,\backslash \{g,h\backslash \}\backslash \}\; +\; \backslash \{g,\backslash \{h,f\backslash \}\backslash \}\; +\; \backslash \{h,\backslash \{f,g\backslash \}\backslash \}\; =\; 0$$

follows from the corresponding identity for the Lie bracket of vector fields, but this is true only up to a locally constant function. However, to prove the Jacobi identity for the Poisson bracket, it is sufficient to show that:

$$\backslash operatorname\{ad\}\_\{\backslash \{g,f\backslash \}\}=\backslash operatorname\{ad\}\_\{-\backslash \{f,g\backslash \}\}=[\backslash operatorname\{ad\}\_f,\backslash operatorname\{ad\}\_g]$$

where the operator $\backslash operatorname\{ad\}\_g$ on smooth functions on {{math|M}} is defined by $\backslash operatorname\{ad\}\_g(\backslash cdot)\; \backslash ;=\backslash ;\; \backslash \{\backslash cdot,\backslash ,\; g\backslash \}$ and the bracket on the right-hand side is the commutator of operators, $[\backslash operatorname\; A,\backslash ,\; \backslash operatorname\; B]\; \backslash ;=\backslash ;\; \backslash operatorname\; A\backslash operatorname\; B\; -\; \backslash operatorname\; B\backslash operatorname\; A$. By {{EquationNote|1|(1)}}, the operator $\backslash operatorname\{ad\}\_g$ is equal to the operator {{math|X_g}}. The proof of the Jacobi identity follows from {{EquationNote|3|(3)}} because, up to the factor of -1, the Lie bracket of vector fields is just their commutator as differential operators.

The algebra of smooth functions on M, together with the Poisson bracket forms a Poisson algebra, because it is a Lie algebra under the Poisson bracket, which additionally satisfies Leibniz's rule {{EquationNote|2|(2)}}. We have shown that every symplectic manifold is a Poisson manifold, that is a manifold with a "curly-bracket" operator on smooth functions such that the smooth functions form a Poisson algebra. However, not every Poisson manifold arises in this way, because Poisson manifolds allow for degeneracy which cannot arise in the symplectic case.

Given a smooth vector field $X$ on the configuration space, let $P\_X$ be its conjugate momentum. The conjugate momentum mapping is a Lie algebra anti-homomorphism from the Lie bracket to the Poisson bracket:

$$\backslash \{P\_X,\; P\_Y\backslash \}\; =\; -P\_\{[X,\; Y]\}.$$

This important result is worth a short proof. Write a vector field $X$ at point $q$ in the configuration space as

$$X\_q\; =\; \backslash sum\_i\; X^i(q)\; \backslash frac\{\backslash partial\}\{\backslash partial\; q^i\}$$

where $\backslash frac\{\backslash partial\}\{\backslash partial\; q^i\}$ is the local coordinate frame. The conjugate momentum to $X$ has the expression

$$P\_X(q,\; p)\; =\; \backslash sum\_i\; X^i(q)\; \backslash ;p\_i$$

where the $p\_i$ are the momentum functions conjugate to the coordinates. One then has, for a point $(q,p)$ in the phase space,

$$\backslash begin\{align\}$$

\{P_X,P_Y\}(q,p) &= \sum_i \sum_j \left\{ X^i(q) \;p_i, Y^j(q)\; p_j \right\} \\

&= \sum_{ij} p_i Y^j(q) \frac{\partial X^i}{\partial q^j} - p_j X^i(q) \frac{\partial Y^j}{\partial q^i} \\

&= -\sum_i p_i \; [X, Y]^i(q) \\

&= - P_{[X, Y]}(q, p).

\end{align}

The above holds for all $(q,\; p)$, giving the desired result.

Poisson brackets deform to Moyal brackets upon quantization, that is, they generalize to a different Lie algebra, the Moyal algebra, or, equivalently in Hilbert space, quantum commutators. The Wigner-İnönü group contraction of these (the classical limit, {{math|ħ → 0}}) yields the above Lie algebra.

To state this more explicitly and precisely, the universal enveloping algebra of the Heisenberg algebra is the Weyl algebra (modulo the relation that the center be the unit). The Moyal product is then a special case of the star product on the algebra of symbols. An explicit definition of the algebra of symbols, and the star product is given in the article on the universal enveloping algebra.

{{colbegin}}

• Commutator
• Dirac bracket
• Lagrange bracket
• Moyal bracket
• Peierls bracket
• Phase space
• Poisson algebra
• Poisson ring
• Poisson superalgebra
• Poisson superbracket

{{colend}}

{{reflist|group=Note}}

{{reflist}}

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